sets of uniqueness and sets of multiplicity
... series not vanishing identically and converging to zero outside E. If no such series exists, E is said to be a set of uniqueness. Every set of positive measure is a set of multiplicity, but there exist sets of multiplicity of measure zero. On the other hand, among perfect sets of measure zero there ...
... series not vanishing identically and converging to zero outside E. If no such series exists, E is said to be a set of uniqueness. Every set of positive measure is a set of multiplicity, but there exist sets of multiplicity of measure zero. On the other hand, among perfect sets of measure zero there ...
QUADRATIC RESIDUES (MA2316, FOURTH WEEK) An integer a is
... Proof. Let us consider the factorisation xp−1 − 1 = (x(p−1)/2 − 1)(x(p−1)/2 + 1). The roots of the left hand side are all nonzero elements modulo p, and each quadratic residue is manifestly a root of the first factor on the right. Since there are p−1 2 quadratic residues, and a polynomial of degree ...
... Proof. Let us consider the factorisation xp−1 − 1 = (x(p−1)/2 − 1)(x(p−1)/2 + 1). The roots of the left hand side are all nonzero elements modulo p, and each quadratic residue is manifestly a root of the first factor on the right. Since there are p−1 2 quadratic residues, and a polynomial of degree ...
CHAP10 Ordinal and Cardinal Numbers
... Then there is some X ⊆ β which is transitive but not an ordinal. Hence X is not well-ordered by ∈. Let 0 ≠ Y ⊆ X and suppose that it has no least. Now the elements of Y are elements of the ordinal β and so are ordinals, and hence transitive. Hence ∩Y is transitive. Let y ∈ Y. Then ∩Y ⊆ y ∈ Y ⊆ X ⊆ β ...
... Then there is some X ⊆ β which is transitive but not an ordinal. Hence X is not well-ordered by ∈. Let 0 ≠ Y ⊆ X and suppose that it has no least. Now the elements of Y are elements of the ordinal β and so are ordinals, and hence transitive. Hence ∩Y is transitive. Let y ∈ Y. Then ∩Y ⊆ y ∈ Y ⊆ X ⊆ β ...
POSriTVE DEFINITE MATRICES AND CATALAN NUMBERS
... example). Joan Birman asked the second author whether it is always possible to make all the elements on the sub and super diagonals one, provided that det(v4) = 1. It is readily shown that this is not possible. For example, if A is positive definite and is not integrally congruent to the identity, t ...
... example). Joan Birman asked the second author whether it is always possible to make all the elements on the sub and super diagonals one, provided that det(v4) = 1. It is readily shown that this is not possible. For example, if A is positive definite and is not integrally congruent to the identity, t ...
Irrationality of the Zeta Constants
... for example, the number 6/π = p≥2 (1 − p−2 ). An estimate of the partial sum of the Dedekind zeta function of quadratic numbers fields will be utilized to develop a general technique for proving the irrationality of the zeta constants ζ(2n + 1) from the known irrationality of the beta constants L(2n ...
... for example, the number 6/π = p≥2 (1 − p−2 ). An estimate of the partial sum of the Dedekind zeta function of quadratic numbers fields will be utilized to develop a general technique for proving the irrationality of the zeta constants ζ(2n + 1) from the known irrationality of the beta constants L(2n ...
Bounded negativity of Shimura curves
... Intersection numbers of Shimura curves are known to appear as coefficients of modular forms and coefficients of modular forms are known to grow. This, however, does not directly give a method to prove Theorem 0.1, since in these modularity statements ([HZ76], [Kud78]) the Shimura curves are packaged ...
... Intersection numbers of Shimura curves are known to appear as coefficients of modular forms and coefficients of modular forms are known to grow. This, however, does not directly give a method to prove Theorem 0.1, since in these modularity statements ([HZ76], [Kud78]) the Shimura curves are packaged ...
Notes on Lecture 3 - People @ EECS at UC Berkeley
... which means that we get a remainder of 1 when dividing N by p1 and by Lemma 3 this is the only possible remainder, while if N were divisible by p1 the remainder would be zero. So N is not divisible by p1 . By the same reasoning, N is not divisible by p2 , nor by p3 , . . . , nor by pk , and we have ...
... which means that we get a remainder of 1 when dividing N by p1 and by Lemma 3 this is the only possible remainder, while if N were divisible by p1 the remainder would be zero. So N is not divisible by p1 . By the same reasoning, N is not divisible by p2 , nor by p3 , . . . , nor by pk , and we have ...
Integer Factorization
... For the algorithms in the first class the run time depends mainly on the size of n, the number being factored, and is not strongly dependent on the size of the factor found. In the second class the run time depends mainly on the size of f , the factor found; cf. [7]. Suppose that we want to find a p ...
... For the algorithms in the first class the run time depends mainly on the size of n, the number being factored, and is not strongly dependent on the size of the factor found. In the second class the run time depends mainly on the size of f , the factor found; cf. [7]. Suppose that we want to find a p ...
Document
... Make a table and look for a pattern. Notice the pattern in how the number of connections ___________. You can use the pattern to make a conjecture. ...
... Make a table and look for a pattern. Notice the pattern in how the number of connections ___________. You can use the pattern to make a conjecture. ...
Prime Numbers
... • We want to determine wether a given large number is prime. • There is no simple yet means of accomplishing this task. • traditionally sieve using trial division – ie. divide by all numbers (primes) in turn less than the square root of the number – only works for small numbers ...
... • We want to determine wether a given large number is prime. • There is no simple yet means of accomplishing this task. • traditionally sieve using trial division – ie. divide by all numbers (primes) in turn less than the square root of the number – only works for small numbers ...
Proving the uncountability of the number of irrational powers of
... is the rational n . irrational. Furthermore, their power composition ( p) An interesting question is how many irrationals α and β exist such that αβ is rational. Lord’s result implies that there are (at least) a countable set of numbers α and β such that αβ is a rational. A countable set is a set wi ...
... is the rational n . irrational. Furthermore, their power composition ( p) An interesting question is how many irrationals α and β exist such that αβ is rational. Lord’s result implies that there are (at least) a countable set of numbers α and β such that αβ is a rational. A countable set is a set wi ...
Prime numbers
... Prime numbers The classical Greek mathematicians were more interested in geometry than in what we would call algebra. (One notable exception was Diophantus, whose surviving work is concerned with integer solutions of polynomial equations, and who was the first Western mathematician to develop any so ...
... Prime numbers The classical Greek mathematicians were more interested in geometry than in what we would call algebra. (One notable exception was Diophantus, whose surviving work is concerned with integer solutions of polynomial equations, and who was the first Western mathematician to develop any so ...
1.4 Deductive Reasoning
... Inductive*reasoning*is*not*a*proof*of*anything*except*for*possibilities*that*you*tested.* There*could*always*be*a*counterexample*just*around*the*corner.* ...
... Inductive*reasoning*is*not*a*proof*of*anything*except*for*possibilities*that*you*tested.* There*could*always*be*a*counterexample*just*around*the*corner.* ...
A formally verified proof of the prime number theorem
... The reason it is useful is that there are two terms in the sum on the left, each sensitive to the presence of primes in different ways. Selberg’s proof involves cleverly balancing the two terms off each other, to show that in the long run, the density of the primes has the appropriate ...
... The reason it is useful is that there are two terms in the sum on the left, each sensitive to the presence of primes in different ways. Selberg’s proof involves cleverly balancing the two terms off each other, to show that in the long run, the density of the primes has the appropriate ...
Elementary primality talk - Dartmouth Math Home
... “The problem of distinguishing prime numbers from composite numbers, and of resolving the latter into their prime factors, is known to be one of the most important and useful in arithmetic. It has engaged the industry and wisdom of ancient and modern geometers to such an extent that it would be supe ...
... “The problem of distinguishing prime numbers from composite numbers, and of resolving the latter into their prime factors, is known to be one of the most important and useful in arithmetic. It has engaged the industry and wisdom of ancient and modern geometers to such an extent that it would be supe ...
TRANSCENDENTAL NUMBERS
... all that I have written here has been in my head for almost a year and it is not in my interest to make a mistake so that one could suspect me of having announced theorems of which I did not have the complete proof. You will publicly ask Jacobi or Gauss to give their opinion not on the truth but on ...
... all that I have written here has been in my head for almost a year and it is not in my interest to make a mistake so that one could suspect me of having announced theorems of which I did not have the complete proof. You will publicly ask Jacobi or Gauss to give their opinion not on the truth but on ...
Predicate Calculus - SIUE Computer Science
... Predicate Calculus Some of the statements that are important in mathematics and computer science are not propositions. For example, X % 2 == 0 is not true for all integers, but only even integers. In order to make statements about the truth of this type of statement, it is necessary to introduce the ...
... Predicate Calculus Some of the statements that are important in mathematics and computer science are not propositions. For example, X % 2 == 0 is not true for all integers, but only even integers. In order to make statements about the truth of this type of statement, it is necessary to introduce the ...
PIANO TUNING AND CONTINUED FRACTIONS 1. Introduction
... A.D. when solving linear equations with infinitely many solutions. However, his use of these interesting mathematical expressions was limited to the specific problems he was solving. Furthermore, ancient Greek and Arab mathematical documents are covered with instances of continued fractions. However ...
... A.D. when solving linear equations with infinitely many solutions. However, his use of these interesting mathematical expressions was limited to the specific problems he was solving. Furthermore, ancient Greek and Arab mathematical documents are covered with instances of continued fractions. However ...
the well-ordering principle - University of Chicago Math
... Proof. Suppose that A has no smallest element; then we have to show that A is empty. We prove the following by induction on n: for all n ∈ N, 1, 2, . . . , n are all not in A. Base case. When n = 1, we have to show that 1 is not in A. But if 1 were in A then it would be the smallest element of A, si ...
... Proof. Suppose that A has no smallest element; then we have to show that A is empty. We prove the following by induction on n: for all n ∈ N, 1, 2, . . . , n are all not in A. Base case. When n = 1, we have to show that 1 is not in A. But if 1 were in A then it would be the smallest element of A, si ...
modulo one uniform distribution of the sequence of logarithms of
... and the sufficiency of Weyl's criterion proves the sequence {y.}°° to be uniformly distributed mod 1. Lemma 2. If a is a positive algebraic number not equal to one, then In a is irrational. Proof. Assume, to the contrary, In a = (p/q), where p and q are non-zero integers. Then e p / q = a9 so that e ...
... and the sufficiency of Weyl's criterion proves the sequence {y.}°° to be uniformly distributed mod 1. Lemma 2. If a is a positive algebraic number not equal to one, then In a is irrational. Proof. Assume, to the contrary, In a = (p/q), where p and q are non-zero integers. Then e p / q = a9 so that e ...